# Typo in Soundararajan's *Bulletin of the AMS* article on Tao's resolution of The Erdos Discrepancy Problem?

I think that there is a typo in the paper referred to in the title of the question, available here. While discussing Roth's proof of AP discrepancy, on page 2, the author states that

Roth [20] established the existence of such irregularities of distribution. Indeed, more generally he showed that if $$A$$ is a subset of the [positive] integers up to $$N$$ with $$|A| = ρN$$, then there exists an absolute positive constant $$c$$ such that $$\sup_{\stackrel{k \leq \sqrt{N}}{a\pmod k}} \left| \sum_{\stackrel{n \in A}{n \equiv a \pmod k}} 1-\frac{\rho N}{q} \right| \geq c \sqrt{\rho(1-\rho)}N^{1/4}$$ In other words, the only subsets of $$[1, N ]$$ that are evenly distributed in all arithmetic progressions with moduli below $$\sqrt{N}$$ are essentially the empty set (with $$\rho=1$$) or the whole set (with $$\rho=1$$).

Shouldn't the $$k$$'s in this inequality should be replaced by $$q$$'s?

• Yes. Or the $q$ should be replaced by $k$. – Thomas Bloom Dec 16 '18 at 9:21
• @ThomasBloom, I will accept this if you want to type it as an answer. Thanks. – kodlu Dec 16 '18 at 23:00

I actually think there is a more significant flaw. A counterexample to the centered inequality is $$\rho = \frac{1}{2}$$ and $$A = \{1,2,\dots,\frac{N}{2}\}$$. Indeed, we always have $$\sum_{n \in A, n \equiv a \pmod{q}} 1 -\frac{N/2}{q} = O(1)$$.
What he actually meant to say is what he ended up saying in words: "evenly distributed in all arithmetic progressions with moduli below $$\sqrt{N}$$". The correct inequality would be something like $$\sup_{\substack{q \le \sqrt{N} \\ a \pmod{q}} \\ 1 \le m_1 < m_2 \le N} \left| \sum_{\substack{n \in A \\ n \equiv a \pmod{q} \\ m_1 < n \le m_2}} 1 - \frac{\rho (m_2-m_1)}{q}\right| \ge c\sqrt{\rho(1-\rho)}N^{1/4}.$$
And indeed, the cited paper says this same thing (note that, by the triangle inequality, we can remove the $$m_1$$ in the above at the cost of changing $$c$$ to $$\frac{c}{2}$$).